Chuse a sequence of rational numbers which converges to a.
Suppose that the function f : R → R is continuous and that f (x) = 0 if x is rational. Prove that f (x) = 0 for all x in R
Now, I know that if f is continuous at x =a and {xn} is a sequence that converges to a, then { f (xn)} converges to f (a). But where does the rational part of the question come into play?
The below is more just a note of passing interest.
Yes! The above is totally correct that is because if we define and by then the continuity of easily implies the continuity of . Thus, one notes that since is Hausdorff that is closed and and so is closed. So, if is dense then and so . But! By definition this means that or that !
Maybe more intuitive is to notice that is a topological group (with the usual addition and usual topology). And since given a topological group the map given by g,h)\mapsto gh^{-1}" alt="\alphag,h)\mapsto gh^{-1}" /> is continuous. Thus, one notes that is continuous and thus so is with . So, noticing that and thus the above claim is true if is closed in . And since is closed in the conclusion follows!
(notice that the above paragraph is a farce since )
No need for a cannon when a flyswatter works.
Theorem: If a continuous function is either positive or negative at then has the same sign on some open interval
So if then is either positive or negative .
But in any interval about there is a rational number.
That contradicts the theorem. HOW?
This is a fascinating proof. Its so algebraic :] I'm not sure I understand it all! I had thought of this low-tech one: suppose D is dense in X and there is a point a at which f and g differ. Use the Hausdorff property to pick disjoint open sets U and V containing f(a) and g(a) respectively. Then is open in X so it meets D, at d say, and f(d)=g(d). But then f(d) is in both U and V, a contradiction.
Here is something I have wondered before but been unable to prove: can we relax the condition that Y is T2? Are there distinct functions agreeing on a dense set that take values in a (say) T1 space? Do you know?
Hmm...
My first inclination would be something like this.
Let be with the cofinite topology and define by . Then, if is open and thus infinite then is infinite and thus open. Similarly, define to be the identity map. Then, they agree on which is dense since given any point and any neighborhood of that neighborhood must infinitely many points and thus a point different from both and .
I hope I didn't make any stupid mistakes.